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3 years ago

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One cubic meter of nitrogen at 40°C and 340kPa is compressed isoentropically to 0.2m^3. Calculate the final pressure when the ni

trogen is reduced to that volume. What is the modulus of elasticity at the beginning and end of the compression? (k = 1.4)

1 answer:

jeka57 [31]3 years ago

8 0

Answer:

a)P₂=3236.21 KPa

b)E=3620.26 KPa

Explanation:

Given that

V₁= 1 m³

T₁=40°C

P₁= 340 KPa

V₂=0.2 m³

k= 1.4

We know that for isentropic process

$PV^{k}=C$

$P_1V_1^{k}=P_2V_2^{k}$

$340\times 1^{1.4}=P_2\times 0.2^{1.4}$

P₂=3236.21 KPa

Final pressure of gas P₂=3236.21 KPa

We know that modulus of elasticity

$E=\dfrac{dP}{-\dfrac{dV}{V}}$

$E=\dfrac{3236.21-340}{-\dfrac{0.2-1}{1}}$

E=3620.26 KPa

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2) The reversible power output is approximately 5456.93 kW

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1) The given parameters are;

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T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

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From which we have;

s₂ = 6.958 kJ/(kg·K)

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From which we have;

s₃ = 7.434 kJ/(kg·K)

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From which we have;

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s₄ = $s_{f4}$ + x₄× $s_{fg}$ = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, $\dot X_{dest}$ , is given as follows;

$\dot X_{dest}$ = T₀ × $\dot S_{gen}$ = T₀ × $\dot m$ × (s₄ + s₂ - s₁ - s₃)

$\dot X_{dest}$ = T₀ × $\dot W$ ×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ $\dot X_{dest}$ = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, $\dot W_{rev}$ = $\dot W_{}$ + $\dot X_{dest}$ ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

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